It looks like you might have intended to say the roots are 7 + i and 5 - i, judging by the extra space between 7 and i.
The simplest polynomial with these characteristics would be
but seeing as each of the options appears to be a quartic polynomial, I suspect f(x) is also supposed to have only real coefficients. In that case, we need to pair up any complex root with its conjugate to "complete" f(x). We end up with
which appears to most closely resemble the third option. Upon expanding, we see f(x) does indeed have real coefficients:
Answer:
35/4, or 8 3/4 as a mixed number
Step-by-step explanation:
First, change 58 1/3 into an improper fraction by multiplying the whole number and denominator, then adding the numerator. 58 x 3 equals 174, + 1 equals 175. So, 58 1/3 as an improper fraction is 175/3. Next, change 6 2/3 into an improper fraction. 6 times 3 equals 18, plus 2 equals 20. So, it's 20/3. So, here are your two fractions:
175/3 & 20/3
To divide fractions, I like to use a method called Keep Change Flip. Basically, you keep the first fractions the same, then change the sign. The division sign changes into a multiplication symbol. Now, your equation should look like this: 175/3 x 20/3. Next, flip the fraction from 20/3 into 3/20. This is what your equation should look like now: 175/3 x 3/20. Now you can multiply the fractions together. Before you do so, you can cross reduce to make it easier. What is a number that both 3 and 3 can be divided by? The correct answer is 3. 3/3= 1, so the equation is now 175/1 x 1/20. However, you can continue to cross reduce. You can also divide 175 and 20 by 5, so the equation changes into this: 35/1 x 1/20. Multiply numerator by numerator, denominator by denominator. So, the answer is 35/4, or 8 3/4 as a mixed number. Hope this helped!
Step-by-step explanation:
we assume day and night are equally long.
and the mouse will start at night of day 1.
then, on day 2, it will first retreat 2 meters (during daylight) and then advance 14 meters (at night).
so, on every full day, the mouse will effectively advance 12 meters (-2 + 14 = 12).
with a starting value of 14 (the first night).
so, we are getting an arithmetic sequence
an = an-1 + c
c = 12 in our case.
a1 = 14
a2 = a1 + 12 = 14+12 = 26
a3 = a2 + 12 = a1 + 12 + 12 = 14 + 2×12 = 38
an = a1 + (n-1)×12 = 14 + 12(n-1)
at what n will it reach 86 ?
86 = 14 + 12(n-1) = 14 + 12n - 12 = 2 + 12n
84 = 12n
n = 7
so, on the 7th day the mouse will reach the surface (if we truly count the first starting night day 1 - this is up to your teacher, but I would).
